high order linear differential equation homogeneous
TRANSCRIPT
Class Notes 7:
High Order Linear Differential Equation
Homogeneous
MAE 82 – Engineering Mathematics
High Order Differential Equations –
Introduction
)(
0)()()()(
)(
0)()()()(
11
1
10
011
1
1
tGytP
dt
dytP
dt
ydtP
dt
ydtP
tGytP
dt
dytP
dt
ydtP
dt
ydtP
nnn
n
n
n
n
n
nn
n
n
High Order Differential Equations –
General Theory
• An nth order ODE has the general form
• We assume that P0,…, Pn, and G are continuous real-valued functions on some interval I = (, ), and that Pn is nowhere zero on I.
• Dividing by Pn, the ODE becomes
• For an nth order ODE, there are typically n initial conditions:
tGytPdt
dytP
dt
ydtP
dt
ydtP
n
n
nn
n
n
)()()()( 011
1
1
tgytpdt
dytp
dt
ydtp
dt
ydyL
n
n
nn
n
)()()( 011
1
1
)1(
00
)1(
0000 ,,, nn ytyytyyty
Theorem 4.1.1
• Consider the nth order initial value problem
• If the functions p1,…, pn, and g are continuous on an open interval I,
• then there exists exactly one solution y = (t) that satisfies the initial value problem. This solution exists throughout the interval I.
)1(
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)1(
0000
011
1
1
,,,
)()()(
nn
n
n
nn
n
ytyytyyty
tgytpdt
dytp
dt
ydtp
dt
yd
High Order Differential Equations –
Homogeneous Equation
• As with 2nd order case, we begin with homogeneous ODE:
• If y1,…, yn are solns to ODE, then so is linear combination
• Every solution can be expressed in this form, with coefficients determined by initial conditions, if we can solve:
0)()()( 011
1
1
ytpdt
dytp
dt
ydtp
dt
ydyL
n
n
nn
n
)()()()( 2211 tyctyctycty nn
)1(
00
)1(
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)1(
11
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00011
)()(
)()(
)()(
nn
nn
n
nn
nn
ytyctyc
ytyctyc
ytyctyc
High Order Differential Equations –
Homogeneous Equation & Wronskian
• The system of equations on the previous slide has a unique solution if its determinant, or Wronskian, is nonzero at t0:
• Since t0 can be any point in the interval I, the Wronskian determinant needs to be nonzero at every point in I.
• As before, it turns out that the Wronskian is either zero for every point in I, or it is never zero on I.
0
)()()(
)()()(
)()()(
,,,
0
)1(
0
)1(
20
)1(
1
00201
00201
021
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tytyty
tytyty
tyyyW
n
n
nn
n
n
n
Theorem 4.1.3
• If {y1,…, yn} is a fundamental set of solutions of
on an interval I,
• then {y1,…, yn} are linearly independent on that interval.
• if {y1,…, yn} are linearly independent solutions to the above
differential equation,
• then they form a fundamental set of solutions on the interval I
0)()()()( 1
)1(
1
)(
ytpytpytpyyL nn
nn
High Order Differential Equations –
Nonhomogeneous Equation
• Consider the nonhomogeneous equation:
• If Y1, Y2 are solutions to nonhomogeneous equation, then Y1 - Y2 is a solution to the homogeneous equation:
• Then there exist coefficients c1,…, cn such that
• Thus the general solution to the nonhomogeneous ODE is
where Y is any particular solution to nonhomogeneous ODE.
)()()()( 011
1
1 tgytpdt
dytp
dt
ydtp
dt
ydyL
n
n
nn
n
0)()(2121 tgtgYLYLYYL
)()()()()( 221121 tyctyctyctYtY nn
)()()()()( 2211 tYtyctyctycty nn
High Order Differential Equations –
Homogeneous Equation with Constant Coefficients
• Consider the nth order linear homogeneous differential equation with constant, real coefficients:
• As with second order linear equations with constant coefficients, y = ert is a solution for values of r that make characteristic polynomial Z(r) zero:
• By the fundamental theorem of algebra, a polynomial of degree n has n roots r1, r2, …, rn, and hence
001
)1(
1
)(
yayayayayL n
n
n
n
0
)( polynomial sticcharacteri
01
1
1
rZ
n
n
n
n
rtrt arararaeeL
)())(()( 21 nn rrrrrrarZ
High Order Differential Equations –
Homogeneous Equation with Constant Coefficients
Cases
• a0,a1,…, an – Real constants an≠0
• Characteristic Polynomial
• Characteristic Equation
001
)1(
1
)(
yayayayayL n
n
n
n
rtey
01
1
1 arararaeeL n
n
n
n
rtrt
01
1
1)( ararararZ n
n
n
n
0)( rZ
High Order Differential Equations –
Homogeneous Equation with Constant Coefficients
Cases
• Z(r) – {Polynomial of degree n}
n zeros, r1, r2,…, rn
Some of which may be equal
(1) Real and unequal roots
(2) Complex roots
(3) Repeated roots
nn rrrrrrarZ 21)(
Case 1: Real & Unequal Roots
• Real and no two are equal n distinct solutions
• Complimentary Solution (Homogeneous)
tr
n
trtr neCeCeCy 21
21
Case 1: Real & Unequal Roots – Example
• Assume
0674 yyyyy
1)0(
2)0(
0)0(
1)0(
y
y
y
y
067 234 rrrr
rtey
Case 1: Real & Unequal Roots – Example
• Roots
• The solution
I.C
3
2
1
1
4
3
2
1
r
r
r
r
tttt
tttt
tttt
tttt
eCeCeCeCy
eCeCeCeCy
eCeCeCeCy
eCeCeCeCy
3
4
2
321
3
4
2
321
3
4
2
321
3
4
2
321
278
94
32
1278 1)0(
294 2)0(
032 0)0(
1 1)0(
4321
4321
4321
4321
CCCCy
CCCCy
CCCCy
CCCCy
Case 1: Real & Unequal Roots – Example
8
1
3
2
12
5
8
11
4
3
2
1
C
C
C
C
tttt eeeey 32
8
1
3
2
12
5
8
11
• Two complex conjugate roots
• Solution
Case 2: Complex Roots
xixeexyi xi sincos)(
xey
xey
x
x
sin
cos
1
1
xeCxeCxy xx sincos)( 21
Case 2: Complex Roots – Example
2)0(
25)0(
4)0(
27)0(
y
y
y
y
04 yy
Case 2: Complex Roots
tecteccomplexir
ir
ececrealir
ir
rrr
re
eer
ey
tt
trtr
rt
rtrt
rt
sincos10
10
01
01
0)1)(1()1(
0)1(
0
43
4
3
21
2
1
224
4
4
21
ngsubstituti
Case 2: Complex Roots
12
1
3
0
22)0(2
52
5)0(
44)0(2
72
7)0(
sincos
cossin
sincos
cossin
sincos
4
3
2
1
421
321
421
321
4321
)4(
4321
4321
4321
4321
c
c
c
c
cccy
cccy
cccy
cccy
tctcececy
tctcececy
tctcececy
tctcececy
tctcececy
tt
tt
tt
tt
tt
Case 2: Complex Roots
ttey t sincos2
13
tteey
y
y
c
tt sin16
17cos
2
1
32
95
32
1
2)0(
815)0(
01
issolutiontheofinstead
withbutICfirstsametheFor
conditionsinitialtheGivenNote
Case 2: Complex Roots
815)0(2)0(
,321.
yy
t
et
change
largetocoef
smallawithevensolutionthedominate
Case 3: Repeated Roots
xexyxexy
xexyxexy
xxeyxxey
xeyxey
i
xey
xey
xixeexyi
xs
s
xs
s
xx
xx
xx
x
x
xxi
sincos
sincos
sincos
sincos
)(
sin
cos
sincos)(
2212
2
6
2
5
43
21
2
1
)(
timessrepeatedrootsconjugateComplex
0
1
2
s
Case 3: Repeated Roots – Example
i
i
i
i
r
rrrr
yyy
10
10
10
10
0)1)(1(12
02
2224
(4)
)sin()cos()sin()cos(
)sin()cos()sin()cos(
4321
4321
ttcttctctcy
ttecttectectecy tttt
